RBSE Solutions Class 12 Chemistry Chapter 10 Halogen Derivatives

Rajasthan Board RBSE Class 12 Chemistry Chapter 10 Halogen Derivatives

RBSE Class 12 Chemistry Chapter 10 Text Book Questions

RBSE Class 12 Chemistry Chapter 10 Multiple Choice Questions

 

Question 1. Out of the following compounds which compound will give Haloform reaction?
(a) Methanol
(b) Ethanol
(c) 1-Propanol
(d) Butanol
Answer: (b) Ethanol
In simple words: Ethanol is the only compound among these options that has the specific chemical structure (a methyl ketone or a secondary alcohol that can be oxidized to a methyl ketone) required to perform the haloform reaction.

🎯 Exam Tip: Remember that compounds with a \( \text{CH}_3\text{CO-} \) group or a \( \text{CH}_3\text{CH(OH)-} \) group can give a positive haloform test.

 

Question 3. Example of haloarene is:
(a) \( \text{CH}_3\text{Cl} \)
(b) \( \text{C}_6\text{H - CH}_2\text{Cl} \)
(c) \( \text{C}_6\text{H}_6\text{Cl}_6 \)
(d) \( \text{C}_6\text{H}_5\text{Cl} \)
Answer: (d) C6H5Cl
In simple words: A haloarene is a compound where a halogen atom is directly connected to a benzene ring. \( \text{C}_6\text{H}_5\text{Cl} \) (chlorobenzene) fits this description because chlorine is attached directly to the phenyl group.

🎯 Exam Tip: A key characteristic of haloarenes is the direct attachment of the halogen to an aromatic ring, distinguishing them from alkyl halides or benzylic halides.

 

Question 4. Which compound will form a yellow precipitate with \( \text{AgNO}_3 \)?
(a) \( \text{CHI}_3 \)
(b) \( \text{CH}_3\text{I} \)
(c) \( \text{CHCl}_3 \)
(d) \( \text{CH}_3 \text{- CH}_2\text{I} \)
Answer: (a) CHI3
In simple words: When \( \text{CHI}_3 \) (iodoform) reacts with \( \text{AgNO}_3 \), it forms a yellow solid called silver iodide. This is a common test to identify iodine-containing compounds.

🎯 Exam Tip: The formation of a yellow precipitate with silver nitrate is a characteristic test for iodide ions or compounds that can easily release iodide ions.

 

Question 5. The intermediate formed in carbylamine reaction is:
(a) \( \text{CN}^{-} \)
(b) \( \text{N = C} \)
(c) \( \text{CCl}_2 \)
(d) \( \text{Cl}^{-} \)
Answer: (c) CCl2
In simple words: In the carbylamine reaction, a highly reactive molecule called dichlorocarbene \( \text{(CCl}_2) \) is temporarily formed as an intermediate before the final product is made.

🎯 Exam Tip: The carbylamine reaction (also known as the isocyanide test) specifically involves the formation of dichlorocarbene as a key step, leading to the characteristic foul-smelling isocyanide product.

 

Question 6. What forms in SN² mechanism?
(a) Transition state
(b) Carbonium ion
(c) Carbanion
(d) Free radical
Answer: (a) Transition state
In simple words: In an \( \text{SN}^2 \) reaction, the old bond breaks and the new bond forms at the same time through a single, unstable intermediate structure called a transition state. No stable ion is formed.

🎯 Exam Tip: The \( \text{SN}^2 \) mechanism is a concerted one-step process, meaning there are no distinct carbocation or carbanion intermediates, only a fleeting transition state.

RBSE Class 12 Chemistry Chapter 10 Very Short Answer Type Questions

 

Question 1. Write full form or name of DDT and BHC.
Answer:
DDT = Dichloro diphenyl trichloroethane
BHC = Benzene hexachloride
In simple words: DDT stands for a complex chemical name and BHC is short for another chemical name, both of which are common pesticides.

🎯 Exam Tip: When asked for full forms, make sure to write out each component of the acronym correctly and completely.

 

Question 2. Write name and formula of anyone tertiary alkyl halide.
Answer:
Formula: \( \text{CH}_3\text{-C}(\text{CH}_3)(\text{Cl})\text{-CH}_3 \)
Name: Tertiary butyl chloride
In simple words: A tertiary alkyl halide has a halogen atom attached to a carbon atom that is itself connected to three other carbon atoms. Tertiary butyl chloride is an example.

🎯 Exam Tip: To correctly identify a tertiary alkyl halide, locate the carbon atom bonded to the halogen and count how many other carbon atoms it is attached to (it should be three).

 

Question 3. Write the name and formula of anyone alcohol and one ketone which give haloform reaction:
Answer:
Alcohol = \( \text{CH}_3\text{-CH(OH)-CH}_3 \)
Name: 2-Propanol
Ketone: \( \text{CH}_3\text{-C}(=\text{O})\text{-CH}_3 \)
Name: 2-Propanone
In simple words: Both 2-propanol (an alcohol) and 2-propanone (a ketone) have a specific methyl group attached to either a carbon with an alcohol group or a carbonyl group, which allows them to react in the haloform reaction.

🎯 Exam Tip: Remember that alcohols like 2-propanol can be oxidized to ketones like 2-propanone, and both functional groups can undergo the haloform reaction if they contain the \( \text{CH}_3\text{CH(OH)-} \) or \( \text{CH}_3\text{CO-} \) moiety.

 

Question 5. CH3-CH-CH-CH3. Br Cl Write the IUPAC name of
Answer:
The given structure is: \( \text{CH}_3\text{-CH(Br)-CH(Cl)-CH}_3 \)
IUPAC name: 2-Bromo-3-chlorobutane
In simple words: To name this molecule, you count the carbons in the longest chain (which is four carbons long, so butane). Then, you place the bromine and chlorine atoms at their correct positions, remembering to list them alphabetically.

🎯 Exam Tip: When naming compounds with multiple halogen substituents, always number the carbon chain to give the substituents the lowest possible numbers and list them in alphabetical order.

 

Question 6. Give any three examples of electrophiles and nucleophiles?
Answer:
Electrophiles: \( \text{H}^{+}, \text{NO}_2^{+}, \text{Cl}^{+} \)
Nucleophiles: \( \text{OH}^{-}, \text{OR}^{-}, \text{Cl}^{-} \)
In simple words: Electrophiles are "electron-loving" species that are attracted to negative charges, so they are often positive or have empty orbitals. Nucleophiles are "nucleus-loving" species that are attracted to positive charges, so they have extra electrons or negative charges.

🎯 Exam Tip: Electrophiles are typically electron-deficient, while nucleophiles are electron-rich. Strong acids often act as electrophiles, and species with lone pairs or negative charges act as nucleophiles.

 

Question 7. Which compound is used in a fire extinguisher?
Answer:
Carbon tetrachloride \( \text{(CCl}_4) \) or tetrachloromethane is used in the fire extinguisher.
In simple words: Carbon tetrachloride is a chemical that was commonly used in fire extinguishers because it does not burn and can help put out fires.

🎯 Exam Tip: Note that while carbon tetrachloride was historically used, its use has largely been phased out due to environmental and health concerns.

 

Question 8. Write the formula of DDT and BHC.
Answer:
DDT: \( \text{C}_{14}\text{H}_9\text{Cl}_5 \)
BHC (Benzene hexachloride): \( \text{C}_6\text{H}_6\text{Cl}_6 \)
In simple words: These are the chemical formulas for DDT and BHC, two important organic compounds used as pesticides.

🎯 Exam Tip: When asked for a "formula," if a structural formula is not explicitly requested or provided in the context, a molecular formula is often acceptable. For BHC, the structural formula is a cyclohexane ring with 6 chlorine atoms attached.

 

Question 9. Write the possible dichloro derivatives of propane.
Answer:
1,3-Dichloropropane: \( \text{Cl-CH}_2\text{-CH}_2\text{-CH}_2\text{-Cl} \)
1,2-Dichloropropane: \( \text{Cl-CH}_2\text{-CH(Cl)-CH}_3 \)
1,1-Dichloropropane: \( \text{CH}_3\text{-CH}_2\text{-CHCl}_2 \)
2,2-Dichloropropane: \( \text{CH}_3\text{-C(Cl)}_2\text{-CH}_3 \)
In simple words: Propane is a three-carbon chain. You can attach two chlorine atoms in different ways to this chain, leading to four possible compounds, each with a unique structure and name.

🎯 Exam Tip: When identifying isomers, systematically consider all possible positions for the substituents on the carbon chain to ensure no isomer is missed.

 

Question 10. Write Hunsdiecker reaction.
Answer:
Hunsdiecker Reaction: A silver salt of a fatty acid can be converted into a bromoalkane by reacting it with bromine in the presence of carbon tetrachloride. This reaction is also known as Borodine Hunsdiecker reaction.
Example Reaction:
\[ \text{R-COOAg} + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{R-Br} + \text{CO}_2 \uparrow + \text{AgBr} \downarrow \]
For example, silver benzoate reacts with bromine to form bromobenzene.
\[ \text{C}_6\text{H}_5\text{COOAg (Silver benzoate)} + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{C}_6\text{H}_5\text{Br (Bromobenzene)} + \text{CO}_2 + \text{AgBr} \]
In simple words: The Hunsdiecker reaction is a way to change a silver salt of a carboxylic acid into an alkyl or aryl bromide. In this process, the acid loses one carbon atom as carbon dioxide gas.

🎯 Exam Tip: Remember the specific reagents and conditions for the Hunsdiecker reaction: bromine in carbon tetrachloride solvent, and a silver carboxylate salt. The reaction involves decarboxylation.

 

Question 12. Which is the best reagent to obtain pure chloroform?
Answer:
Chloral hydrate is the best reagent to obtain pure chloroform. The reaction is:
\[ \text{CCl}_3\text{-CH(OH)}_2 + \text{NaOH} \rightarrow \text{CHCl}_3 + \text{HCOONa} + \text{H}_2\text{O} \]
Chloral hydrate (trichloroacetaldehyde hydrate) reacts with a strong base like sodium hydroxide to yield chloroform.
In simple words: To get very pure chloroform, you should use chloral hydrate and react it with a strong base.

🎯 Exam Tip: Chloral hydrate is a stable compound that readily undergoes hydrolysis in the presence of a base to produce chloroform, making it a convenient source for laboratory preparation.

 

Question 13. Which gas is formed when chloroform is exposed to air?
Answer:
When chloroform \( \text{(CHCl}_3) \) is exposed to air and light, poisonous phosgene gas \( \text{(COCl}_2) \) is formed. This is a very dangerous reaction.
The reaction is:
\[ \text{2CHCl}_3 + \text{O}_2 \xrightarrow{\text{Light}} \text{2COCl}_2 + \text{2HCl} \]
In simple words: If chloroform is left out in the open air and light, it slowly changes into a very toxic gas called phosgene, which is harmful.

🎯 Exam Tip: Chloroform must be stored in dark, amber-colored bottles filled to the brim to prevent its oxidation into phosgene, which is a highly toxic gas.

 

Question 14. Which one is more reactive from methyl chloride and methyl iodide?
Answer:
Methyl iodide \( \text{(CH}_3\text{I)} \) will be more reactive than methyl chloride \( \text{(CH}_3\text{Cl)} \). This is because the iodide ion \( \text{(I}^-) \) is a better leaving group than the chloride ion \( \text{(Cl}^-) \). Also, the carbon-iodine bond (C-I) is weaker and has a lower bond dissociation enthalpy compared to the carbon-chlorine bond (C-Cl).
In simple words: Methyl iodide reacts faster than methyl chloride because the iodine atom is larger and leaves more easily from the carbon, and the bond between carbon and iodine is weaker.

🎯 Exam Tip: In nucleophilic substitution reactions, the reactivity of alkyl halides often depends on the leaving group ability, which generally increases with increasing size of the halogen atom (I > Br > Cl > F).

 

Question 16. What is the use of DDT?
Answer:
DDT is used as an insecticide for killing or controlling mosquitoes, lice, and other pests. It was widely used in the past to prevent diseases like malaria.
In simple words: DDT is a chemical that kills insects like mosquitoes and lice.

🎯 Exam Tip: While DDT was effective as a pesticide, its persistence in the environment and harmful effects on wildlife led to its ban in many countries.

 

Question 17. Write two examples of 2° alkyl halide.
Answer:
1. 2-Chloropropane: \( \text{CH}_3\text{-CH(Cl)-CH}_3 \)
2. 2-Iodobutane: \( \text{CH}_3\text{-CH(I)-CH}_2\text{-CH}_3 \)
In simple words: A secondary alkyl halide has a halogen atom attached to a carbon atom that is connected to two other carbon atoms. 2-chloropropane and 2-iodobutane are examples.

🎯 Exam Tip: To identify a secondary alkyl halide, check that the carbon atom bearing the halogen is bonded to exactly two other carbon atoms and one hydrogen atom.

 

Question 18. Arrange the following in order of reactivity towards SN1 reaction.
Answer:
The order of reactivity towards \( \text{SN}^1 \) reaction depends on the stability of the carbocation formed. Tertiary carbocations are more stable than secondary, which are more stable than primary. Therefore, the order is:
\[ \text{CH}_3\text{-Br} < \text{CH}_3\text{-CH(CH}_3)\text{-Br} < \text{CH}_3\text{-C(CH}_3)_2\text{-Br} \]
(1° alkyl halide) < (2° alkyl halide) < (3° alkyl halide)
In simple words: For \( \text{SN}^1 \) reactions, alkyl halides that can form more stable carbocations (like tertiary ones) react faster than those that form less stable ones (like primary ones).

🎯 Exam Tip: Always consider the stability of the carbocation intermediate when determining the reactivity order for \( \text{SN}^1 \) reactions; tertiary carbocations are the most stable due to hyperconjugation and inductive effects.

RBSE Class 12 Chemistry Chapter 10 Short Answer Type Questions

 

Question 1. \( \text{C}_6\text{H}_5\text{Cl} \) is less reactive towards nucleophilic substitution reactions than \( \text{C}_2\text{H}_5\text{Cl} \). Explain.
Answer:
In chloroethane \( \text{(C}_2\text{H}_5\text{Cl)} \), the carbon atom linked to the chlorine atom is \( \text{sp}^3 \) hybridized. In chlorobenzene \( \text{(C}_6\text{H}_5\text{Cl)} \), the carbon atom linked to the chlorine atom is \( \text{sp}^2 \) hybridized. An \( \text{sp}^2 \) hybridized carbon atom has more s-character, making it more electronegative. This allows it to hold the electron pair of the carbon-chlorine bond more tightly compared to an \( \text{sp}^3 \) hybridized carbon. This makes the carbon-chlorine bond cleavage in chlorobenzene more difficult, leading to its lower reactivity compared to chloroethane in nucleophilic substitution reactions.
In simple words: Chlorobenzene reacts slower than chloroethane in nucleophilic substitution because the carbon atom bonded to chlorine in chlorobenzene holds onto the bond more strongly due to its sp2 hybridization.

🎯 Exam Tip: The partial double bond character of the C-Cl bond in chlorobenzene due to resonance, along with the higher electronegativity of the \( \text{sp}^2 \) hybridized carbon, significantly reduces its reactivity in nucleophilic substitution reactions.

 

Question 3. Write the chemical reaction for the formation of BHC.
Answer:
Benzene hexachloride (BHC) is formed when benzene reacts with three molecules of chlorine in the presence of ultraviolet (UV) light.
\[ \text{C}_6\text{H}_6 + \text{3Cl}_2 \xrightarrow{\text{hv (UV Light)}} \text{C}_6\text{H}_6\text{Cl}_6 \]
The reaction proceeds via a free radical addition mechanism.
In simple words: BHC is made by mixing benzene with chlorine gas and shining UV light on it, causing chlorine atoms to add to the benzene ring.

🎯 Exam Tip: The presence of UV light is crucial for this addition reaction; without it, benzene typically undergoes substitution reactions with halogens in the presence of a Lewis acid catalyst.

 

Question 4. How will you prepare the following compounds from chlorobenzene:
(a) Phenol
(b) Diphenyl
Answer:
(a) Phenol from chlorobenzene:
Phenol can be prepared from chlorobenzene through the Dow process. Chlorobenzene is heated with aqueous sodium hydroxide under high pressure and temperature to form sodium phenoxide, which is then acidified to yield phenol.
\[ \text{C}_6\text{H}_5\text{Cl} + \text{2NaOH} \xrightarrow[\text{-NaCl, -H}_2\text{O}]{\text{623 K/300 atm}} \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ \xrightarrow{\text{HCl}} \text{C}_6\text{H}_5\text{OH} \]
(b) Diphenyl from chlorobenzene:
Diphenyl can be prepared from chlorobenzene using the Wurtz-Fittig reaction, where two molecules of chlorobenzene react with sodium in the presence of dry ether.
\[ \text{C}_6\text{H}_5\text{Cl} + \text{2Na} + \text{C}_6\text{H}_5\text{Cl} \xrightarrow{\text{Anhydrous ether}} \text{C}_6\text{H}_5\text{-C}_6\text{H}_5 + \text{2NaCl} \]
In simple words: To make phenol from chlorobenzene, you heat it with strong base and then add acid. To make diphenyl, you react two chlorobenzene molecules with sodium metal.

🎯 Exam Tip: Remember the specific conditions (high temperature/pressure for Dow process, sodium/ether for Wurtz-Fittig) for converting chlorobenzene to different aromatic compounds due to its deactivated nature.

 

Question 5. Explain \( \beta \)-elimination.
Answer:
\( \beta \)-elimination is a reaction where a haloalkane (or alkyl halide) with a hydrogen atom on the carbon adjacent to the halogen-bearing carbon (\( \beta \)-carbon) is heated with an alcoholic solution of potassium hydroxide (alc. KOH). In this reaction, the hydroxide ion \( \text{(OH}^-) \) acts as a base, abstracting a hydrogen atom from the \( \beta \)-carbon. Simultaneously, the halogen atom leaves from the \( \alpha \)-carbon (the carbon bonded to the halogen). This concerted removal of a hydrogen and a halogen atom results in the formation of an alkene.
General Reaction:
\[ \text{R-CH(X)-CH}_2\text{-R'} + \text{alc. KOH} \rightarrow \text{R-CH=CH-R'} + \text{KX} + \text{H}_2\text{O} \]
Example:
\[ \text{CH}_3\text{-CH(Br)-CH}_3 + \text{alc. KOH} \rightarrow \text{CH}_3\text{-CH=CH}_2 + \text{KBr} + \text{H}_2\text{O} \]
(2-Bromopropane) (Propene)
This reaction is also called dehydrohalogenation.
In simple words: \( \beta \)-elimination happens when you remove a hydrogen and a halogen from neighboring carbon atoms in a molecule, resulting in the formation of a double bond.

🎯 Exam Tip: Understand that \( \beta \)-elimination is a common reaction for alkyl halides in the presence of strong bases (like alcoholic KOH), leading to the formation of alkenes and often competing with nucleophilic substitution reactions.

 

Question 6. Write the mechanism of Carbyl amine reaction.
Answer:
The Carbylamine Reaction (also called the Isocyanide Reaction) is a test for primary amines. When a primary amine (aliphatic or aromatic) is heated with chloroform \( \text{(CHCl}_3) \) and alcoholic potassium hydroxide \( \text{(KOH)} \), it forms an isocyanide (or carbylamine). Isocyanides have a very unpleasant smell.
The reaction involves the generation of dichlorocarbene \( \text{(CCl}_2) \) as an intermediate.
General Reaction:
\[ \text{R-NH}_2 + \text{CHCl}_3 + \text{3KOH (alc)} \rightarrow \text{R-NC} + \text{3KCl} + \text{3H}_2\text{O} \]
Example with methylamine:
\[ \text{CH}_3\text{-NH}_2 \text{ (Methylamine)} + \text{CHCl}_3 + \text{3KOH (alc)} \rightarrow \text{CH}_3\text{-NC (Methyl isocyanide)} + \text{3KCl} + \text{3H}_2\text{O} \]
Example with aniline:
\[ \text{C}_6\text{H}_5\text{NH}_2 \text{ (Aniline)} + \text{CHCl}_3 + \text{3KOH (alc)} \rightarrow \text{C}_6\text{H}_5\text{NC (Phenyl isocyanide)} + \text{3KCl} + \text{3H}_2\text{O} \]
This reaction is not given by secondary or tertiary amines.
In simple words: The carbylamine reaction is a chemical test for primary amines where a bad-smelling substance called isocyanide is formed by reacting the amine with chloroform and a strong base.

🎯 Exam Tip: The distinctive foul odor of the isocyanide product is the key indicator for a positive carbylamine test, confirming the presence of a primary amine.

 

Question 7. How will you prepare the following from chloroform:
(a) Acetylene
(b) \( \text{CCl}_4 \)
(c) Salicylaldehyde
Answer:
(a) Acetylene from Chloroform:
Acetylene can be prepared by warming chloroform with silver powder. The reaction involves the removal of chlorine atoms and formation of a triple bond.
\[ \text{2CHCl}_3 + \text{6Ag} \xrightarrow{\text{Warm}} \text{HC}\equiv\text{CH} + \text{6AgCl} \]
(b) \( \text{CCl}_4 \) from Chloroform:
Carbon tetrachloride \( \text{(CCl}_4) \) is obtained by further chlorination of chloroform in the presence of sunlight.
\[ \text{CHCl}_3 + \text{Cl}_2 \xrightarrow{\text{hv (Sunlight)}} \text{CCl}_4 + \text{HCl} \]
(c) Salicylaldehyde from chloroform:
Salicylaldehyde is prepared from chloroform and phenol in the presence of alcoholic KOH through the Reimer-Tiemann reaction.
\[ \text{C}_6\text{H}_5\text{OH (Phenol)} + \text{CHCl}_3 + \text{3KOH} \xrightarrow{\text{Formylation}} \text{o-HOC}_6\text{H}_4\text{CHO (Salicylaldehyde)} + \text{3KCl} + \text{3H}_2\text{O} \]
In simple words: Chloroform can be used to make acetylene by reacting it with silver, carbon tetrachloride by adding more chlorine, and salicylaldehyde by reacting it with phenol in a specific way.

🎯 Exam Tip: Each transformation requires specific reagents and conditions; for example, silver powder for reduction to alkynes, further halogenation for higher substituted halides, and the Reimer-Tiemann reaction for phenol derivatives.

 

Question 12. Benzyl chloride is more reactive than chlorobenzene. Why?
Answer:
Benzyl chloride is more reactive than chlorobenzene towards \( \text{SN}^1 \) reactions because it can easily ionize to form a benzyl carbocation. This benzyl carbocation is stabilized by resonance with the benzene ring. This resonance delocalizes the positive charge over the ring, making the carbocation more stable and easier to form. In contrast, in chlorobenzene, the chlorine atom is directly attached to an \( \text{sp}^2 \) hybridized carbon of the benzene ring. This bond has partial double bond character due to resonance, which makes it stronger and harder to break. Also, ionization of chlorobenzene would lead to a highly unstable phenyl carbocation, which is not resonance stabilized. Therefore, nucleophilic attack on chlorobenzene is difficult.
In simple words: Benzyl chloride reacts faster because it can form a stable, resonance-supported intermediate called a benzyl carbocation. Chlorobenzene reacts slower because its chlorine bond is stronger and its intermediate is not stable.

🎯 Exam Tip: The key difference in reactivity lies in the ability to form a stable carbocation: benzyl carbocations are stabilized by resonance, while phenyl carbocations are highly unstable, making benzyl halides more reactive in \( \text{SN}^1 \) reactions.

RBSE Class 12 Chemistry Chapter 10 Long Answer Type Questions

 

Question 1. Explain the following:
(a) Classification of halogen derivatives
(b) Nature of C-X bond in halogen derivatives
(c) Directive influence of halogen atom in haloarene.
Answer:
(a) Classification of Halogen Derivatives:
In the IUPAC system, halogen derivatives are named as halo-substituted alkanes. The position of the halogen atom determines the numbering, with the halogen-bearing carbon receiving the lowest possible number. If multiple different halogen atoms are present, they are listed alphabetically.
Halogen derivatives can be broadly classified based on:
(i) Number of halogen atoms: They can be mono-, di-, tri-, or tetrahalo compounds depending on how many halogen atoms are present in the molecule. For example, chloromethane (monohaloalkane), dichloromethane (dihaloalkane), trichloromethane (trihaloalkane), tetrachloromethane (tetrahaloalkane).

(ii) Nature of carbon atom to which halogen is attached (for monohaloalkanes):
Primary Alkyl Halides: The halogen atom is attached to a primary carbon atom (a carbon atom bonded to only one other carbon atom).
Example: Chloromethane \( \text{(CH}_3\text{Cl)} \), Chloroethane \( \text{(CH}_3\text{CH}_2\text{Cl)} \), 1-Chloropropane \( \text{(CH}_3\text{CH}_2\text{CH}_2\text{Cl)} \)
Secondary Alkyl Halides: The halogen atom is attached to a secondary carbon atom (a carbon atom bonded to two other carbon atoms).
Example: Isopropyl chloride (2-Chloropropane) \( \text{CH}_3\text{-CH(Cl)-CH}_3 \)
Tertiary Alkyl Halides: The halogen atom is attached to a tertiary carbon atom (a carbon atom bonded to three other carbon atoms).
Example: Tertiary butyl chloride \( \text{CH}_3\text{-C(CH}_3)(\text{Cl})\text{-CH}_3 \)
(iv) Benzylic Halides: These compounds have a halogen atom bonded to an \( \text{sp}^3 \) hybridized carbon atom that is directly next to an aromatic ring. This carbon is known as a benzylic carbon.
Example: Benzyl chloride \( \text{(C}_6\text{H}_5\text{CH}_2\text{Cl)} \)
(v) Vinylic Halides: In these compounds, the halogen atom is bonded directly to an \( \text{sp}^2 \) hybridized carbon atom of a carbon-carbon double bond.
Example: Vinyl chloride \( \text{(CH}_2\text{=CH-Cl)} \)
(vi) Aryl Halides or Haloarenes: In these compounds, the halogen atom is bonded directly to an \( \text{sp}^2 \) hybridized carbon atom of an aromatic ring.
Example: Chlorobenzene \( \text{(C}_6\text{H}_5\text{Cl)} \)
(b) Nature of C-X bond in halogen derivatives:
In haloalkanes, the carbon-halogen bond \( \text{(C-X)} \) is formed by the overlapping of an \( \text{sp}^3 \) hybrid orbital of carbon and a p orbital of the halogen (e.g., \( \text{3p} \) for chlorine).

The C-X bond is covalent, but there is a significant difference in electronegativity between carbon (2.6) and the halogens (e.g., Cl is 3.16). Because of this difference, the C-X bond is polar. The halogen atom carries a partial negative charge \( \text{(\delta}^-) \) and the carbon atom carries a partial positive charge \( \text{(\delta}^+) \). This polarity influences the reactivity of halogen derivatives.
In simple words: Halogen derivatives are grouped by how many halogens they have and where those halogens are attached (like primary, secondary, tertiary, or to a special carbon near a ring). The bond between carbon and halogen is slightly uneven, with the halogen pulling electrons more strongly, making the bond polar.

🎯 Exam Tip: When classifying halogen derivatives, clearly differentiate between alkyl, vinylic, benzylic, and aryl halides, as their bonding characteristics and reactivity differ significantly.

 

Question 2. How will you prepare:
(a) Alkyl halide from alcohol?
(b) Alkyl halide from halogen exchange?
(c) Chloroform from the acetone?
(d) Salicylic acid from carbon tetrachloride.
Answer:
(a) Alkyl halide from alcohol:
(i) By the reaction of alcohol and halogen acid.
\( \text{R - OH + HCl} \xrightarrow { \text{ZnCl}_2 } \text{R - Cl + HCl} \)
\( \text{R - OH + HBr} \xrightarrow { \text{H}_2\text{SO}_4 } \text{R - Br + H}_2\text{O} \)
(ii) By the action of thionyl chloride on alcohol.
\( \text{R - OH + SOCl}_2 \xrightarrow { \text{C}_5\text{H}_5\text{N} } \text{R - Cl + SO}_2 \text{ + HCl} \)
(iii) By the action of phosphorus halides on alcohol.
\( \text{R - OH + PCl}_5 \xrightarrow { \triangle } \text{R - Cl + POCl}_3 \text{ + HCl} \)
\( \text{3R - OH + PCl}_3 \xrightarrow { \triangle } \text{3R - Cl + H}_3\text{PO}_3 \)

(b) Alkyl halide from halogen exchange:
In this method, alkyl halides are prepared from other alkyl halides. For example, by treating alkyl iodides with silver fluoride, alkyl fluorides are formed.
\( \text{R - I + AgF} \rightarrow \text{R - F + AgI} \)
In simple words: Alkyl halides can be made from alcohols by reacting them with acids, thionyl chloride, or phosphorus halides. They can also be made by exchanging one halogen for another in an existing alkyl halide.

🎯 Exam Tip: Remember the specific reagents and conditions required for each type of conversion. For example, anhydrous \( \text{ZnCl}_2 \) is crucial when using HCl with alcohols.

 

Answer:
(c) Chloroform from Acetone:
Bleaching powder reacts with water to form nascent chlorine, which then reacts with acetone. First, acetone reacts with chlorine to form trichloroacetone. This trichloroacetone then hydrolyzes in the presence of calcium hydroxide (from bleaching powder) to produce chloroform and calcium acetate.
\( \text{CaOCl}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + 2[\text{Cl}] \)
\( \text{CH}_3\text{-CO-CH}_3 + 6[\text{Cl}] \rightarrow \text{CCl}_3\text{-CO-CH}_3 + \text{3HCl} \)
\( \text{CCl}_3\text{-CO-CH}_3 + \text{Ca(OH)}_2 \rightarrow 2\text{CHCl}_3 + (\text{CH}_3\text{COO})_2\text{Ca} \)

(d) Salicylic acid from carbon tetrachloride:
Phenol reacts with carbon tetrachloride in the presence of sodium hydroxide to form salicylic acid. This is part of the Reimer-Tiemann reaction where chloroform is used. With carbon tetrachloride, salicylic acid is formed after intermediate steps.
\( \text{Phenol} + \text{CCl}_4 + 4\text{NaOH} \xrightarrow { \triangle } \text{Salicylic Acid} + 4\text{NaCl} + 2\text{H}_2\text{O} \)
In simple words: To make chloroform from acetone, acetone is first chlorinated, then broken down by calcium hydroxide. To make salicylic acid from carbon tetrachloride, phenol is reacted with carbon tetrachloride and sodium hydroxide.

🎯 Exam Tip: For synthesis questions, it's vital to know the starting materials, reagents, and balanced chemical equations for each step. Practice writing out the full reaction sequence.

 

Question 3. Write short notes on the following:
(a) Haloform Reaction
(b) Carbylamine Reaction
(c) Darzen's Reaction
(d) Sandmeyer Reaction.
Answer:
(a) Haloform Reaction:
The haloform reaction occurs when a methyl ketone (a ketone with a \( \text{CH}_3\text{-CO} \) group) or certain alcohols are treated with a halogen (like iodine, bromine, or chlorine) in the presence of a base (like sodium hydroxide). The methyl ketone is oxidized to a sodium salt of an acid with one less carbon atom, and a haloform (like iodoform, \( \text{CHX}_3 \)) is formed. If iodine is used, a yellow precipitate of iodoform is formed, which is a common test for methyl ketones and secondary alcohols with a \( \text{CH}_3\text{-CH(OH)} \) group.
\( \text{CH}_3\text{-CO-R} + 3\text{X}_2 + 4\text{NaOH} \rightarrow \text{R-COONa} + \text{CHX}_3 + 3\text{NaX} + 3\text{H}_2\text{O} \)
(Here, X represents a halogen like I, Br, or Cl.)
In simple words: Haloform reaction is a test for specific alcohols and ketones. When these chemicals react with a halogen and a base, they produce a haloform compound, like iodoform, which can be seen as a yellow solid.

🎯 Exam Tip: Remember that the haloform reaction specifically requires a methyl ketone group \( (\text{CH}_3\text{-CO-}) \) or a secondary alcohol group \( (\text{CH}_3\text{-CH(OH)-}) \) that can be oxidized to a methyl ketone. This is a key identification test in organic chemistry.

 

Answer:
(b) Carbylamine Reaction:
This reaction is also called the Isocyanide Reaction. It involves heating chloroform with a primary amine (aliphatic or aromatic) and an alcoholic potassium hydroxide (KOH) solution. This process forms an isocyanide or carbylamine, which has a very unpleasant smell. Secondary and tertiary amines do not show this reaction, making it a useful test to identify primary amines and chloroform.
\( \text{CH}_3\text{CH}_2\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc)} \rightarrow \text{CH}_3\text{CH}_2\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
(Ethyl amine + Chloroform + Potassium hydroxide \( \rightarrow \) Ethyl isocyanide + Potassium chloride + Water)

(c) Darzen's Reaction:
Darzen's reaction is a method to convert straight-chain primary alcohols into chloro derivatives without any rearrangement. This happens when the alcohol reacts with thionyl chloride (\( \text{SOCl}_2 \)) in the presence of pyridine. This is a good way to make alkyl chlorides because the byproducts, sulfur dioxide (\( \text{SO}_2 \)) and hydrogen chloride (\( \text{HCl} \)), are gases that easily escape, leaving a pure product.
\( \text{R - OH + SOCl}_2 \xrightarrow { \text{Pyridine} } \text{RCl + SO}_2 \text{ + HCl} \)

(d) Sandmeyer Reaction:
In the Sandmeyer reaction, a benzene diazonium chloride salt reacts with cuprous chloride (\( \text{CuCl} \)) or cuprous bromide (\( \text{CuBr} \)) in the presence of the corresponding halogen acid (HCl or HBr). This reaction is used to form haloarenes (aryl halides) like chlorobenzene or bromobenzene from aryl diazonium salts.
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow { \text{CuCl/HCl} } \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \uparrow \)
(Benzene diazonium chloride \( \rightarrow \) Chlorobenzene + Nitrogen gas)
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow { \text{CuBr/HBr} } \text{C}_6\text{H}_5\text{Br} + \text{N}_2 \uparrow \)
(Benzene diazonium chloride \( \rightarrow \) Bromobenzene + Nitrogen gas)
In simple words: Carbylamine reaction tests for primary amines by making a bad-smelling product. Darzen's reaction turns alcohols into clean chloro compounds. Sandmeyer reaction changes diazonium salts into haloarenes using copper salts.

🎯 Exam Tip: Pay attention to the specific catalysts and conditions for each named reaction. For the Sandmeyer reaction, make sure to use the correct copper(I) salt corresponding to the halogen you want to introduce.

 

Question. Explain the mechanism of unimolecular nucleophilic substitution (SN1) and bimolecular nucleophilic substitution (SN2) reactions.
Answer:
**Unimolecular Nucleophilic Substitution (\( \text{S}_{\text{N}}1 \)) Reaction:**
This reaction occurs in two steps, and its rate depends only on the concentration of the alkyl halide.
(ii) **Step I (Rate-determining step):** The carbon-halogen (C-X) bond undergoes heterolytic cleavage, meaning it breaks unevenly. This forms a carbocation (a carbon atom with a positive charge, \( \text{R}^+ \)) as an intermediate and a halide ion \( (\text{X}^-) \). This first step is slow and therefore determines the overall rate of the reaction.
\( \text{R-X} \xrightarrow [ \text{slow} ]{ \text{step} } \text{R}^+ + \text{X}^- \)
(iii) Since only the alkyl halide concentration affects the rate in the first step, the reaction rate is proportional to the concentration of the alkyl halide. This is why it's called a unimolecular reaction.
\( \text{Rate} \propto [\text{R-X}] \)
(iv) **Step II:** In the second step, the carbocation intermediate is quickly attacked by a nucleophile (\( \text{Nu}^- \)) to form the final product (\( \text{R-Nu} \)). This step is much faster than the first step.
\( \text{R}^+ + \text{Nu}^- \xrightarrow { \text{Fast} } \text{R-Nu} \)
(v) The stability of the carbocation formed in Step I directly affects how easily the reaction proceeds. More stable carbocations lead to faster reactions. The order of carbocation stability is: Tertiary \( > \) Secondary \( > \) Primary \( > \) Methyl.
Consequently, the order of reactivity for \( \text{S}_{\text{N}}1 \) reactions (if the halogen atom is the same) follows the stability of the carbocation: Tertiary \( > \) Secondary \( > \) Primary.
(vi) Since the carbocation intermediate is planar, the nucleophile can attack from either side. If the reactant is an optical isomer, this can lead to a racemic mixture (a mix of both enantiomers).

**Bimolecular Nucleophilic Substitution (\( \text{S}_{\text{N}}2 \)) Reaction:**
This type of reaction happens in a single step, and its rate depends on the concentrations of both the alkyl halide and the nucleophile.
(i) This reaction is a concerted process, meaning bond breaking and bond forming happen simultaneously in one step.
(ii) Instead of an intermediate, a single transition state is formed. In this state, the bond to the leaving group is partially broken, and the bond to the attacking nucleophile is partially formed.
(iii) The attacking nucleophile approaches the carbon atom from the side opposite to where the leaving group is attached. This is known as backside attack or rear attack, specifically at an approximately 180° angle from the leaving group.
(iv) In the transition state, both the attacking nucleophile (\( \text{Nu}^- \)) and the leaving group (\( \text{X}^- \)) are partially bonded to the central carbon atom.
(v) The rate of the reaction is determined by the concentrations of both the alkyl halide and the nucleophile. Hence, these are bimolecular nucleophilic substitution reactions.
\( \text{Rate} \propto [\text{R-X}][\text{Nu}^-] \)
(vi) Because of the backside attack, the configuration of the product is inverted compared to the reactant. This is known as "Walden inversion."
(vii) The presence of bulky alkyl groups on the carbon atom bonded to the halogen creates steric hindrance, making it harder for the nucleophile to attack. Therefore, as the number of alkyl groups increases, the reactivity of the alkyl halide towards \( \text{S}_{\text{N}}2 \) mechanisms decreases. If the halogen atom is the same, the order of reactivity for \( \text{S}_{\text{N}}2 \) reactions is: Methyl \( > \) Primary \( > \) Secondary \( > \) Tertiary.
In simple words: \( \text{S}_{\text{N}}1 \) reactions happen in two steps; first, the halogen leaves, then a nucleophile attacks. Its speed depends only on the original molecule. \( \text{S}_{\text{N}}2 \) reactions happen in one step; the nucleophile pushes the halogen out from the back. Its speed depends on both the original molecule and the attacking nucleophile.

🎯 Exam Tip: Clearly differentiate between \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) by focusing on the number of steps, rate dependence, intermediates vs. transition states, stereochemistry (racemization vs. inversion), and the influence of steric hindrance/carbocation stability.

 

Question 5. Write a short note on the following?
(i) Freons
(ii) DDT
(iii) BHC
Answer:
(i) **Freons:**
Freons are chlorofluoro derivatives of methane and ethane. They are typically prepared by reacting hydrogen fluoride (HF) with carbon tetrachloride (\( \text{CCl}_4 \)) or hexachloroethane in the presence of antimony pentachloride (\( \text{SbCl}_5 \)).
For example:
\( \text{CCl}_4 + \text{HF} \xrightarrow { \text{SbCl}_5 } \text{CCl}_3\text{F} + \text{HCl} \) (Freon-11)
\( \text{CCl}_3\text{F} + \text{HF} \xrightarrow { \text{SbCl}_5 } \text{CCl}_2\text{F}_2 + \text{HCl} \) (Freon-12)
\( \text{C}_2\text{Cl}_6 + 2\text{HF} \xrightarrow { \text{SbCl}_5 } \text{C}_2\text{Cl}_4\text{F}_2 + 2\text{HCl} \) (Freon-112)
**Properties of Freons:** They are odorless, volatile liquids. They are highly inert and very stable, even at high temperatures and pressures.
**Uses of Freons:** They are used as inert solvents, refrigerants in refrigerators and air conditioners, and propellants in aerosols.

(ii) **DDT (Dichlorodiphenyltrichloroethane):**
DDT is synthesized by treating a mixture of chloral with chlorobenzene in the presence of concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). It is a white crystalline solid. DDT was widely used as an insecticide for killing or controlling mosquitoes, lice, and other pests. However, it is a harmful poisonous substance that does not easily decompose in the environment.

(iii) **BHC (Benzene Hexachloride):**
BHC has several common names, such as gammexane and lindane. Its IUPAC name is 1,2,3,4,5,6-hexachlorocyclohexane. It is prepared by the chlorination of benzene in the presence of ultraviolet (UV) radiation. BHC is a mixture of several isomers (alpha, beta, gamma, delta, epsilon, sigma). The gamma-isomer ( \( \gamma \)-BHC) is the most potent and active form, making it a very effective insecticide. Compared to other isomers, the \( \gamma \)-isomer is relatively small, which gives it better penetrating power.
In simple words: Freons are stable chemicals used in cooling and sprays. DDT is an old insecticide, now known to be harmful and doesn't break down easily. BHC is another insecticide made from benzene, and its gamma-form is very strong.

🎯 Exam Tip: For each chemical, remember its full form (if applicable), how it's prepared, its key properties, and its main uses or environmental impact. Pay attention to specific reaction conditions like catalysts or radiation.

 

Question 6. Explain the electrophilic and nucleophilic reaction of chlorobenzene.
Answer:
(i) **Electrophilic substitution reactions of chlorobenzene:**
In chlorobenzene, the chlorine atom has a dual effect. Due to its electron-withdrawing inductive effect (-I effect), it reduces the electron density on the benzene ring, making electrophilic attack slower compared to benzene. This is why chlorobenzene is less reactive towards electrophilic substitution. However, the chlorine atom also has a lone pair of electrons that it can donate to the ring through resonance (+M effect), which increases electron density at the ortho and para positions. The +M effect dominates over the -I effect when directing electrophiles, so electrophilic substitution mainly occurs at the ortho and para positions, even if the overall reaction is slower.
Some examples of electrophilic substitution reactions of chlorobenzene are:
* **Sulphonation:** Forms o-Chlorobenzene sulphonic acid and p-chlorobenzene sulphonic acid.
* **Nitration:** Forms o-Nitrochlorobenzene and p-Nitrochlorobenzene.
* **Alkylation (Friedel-Crafts):** Forms o-Chlorotoluene and p-Chlorotoluene.
* **Acetylation (Friedel-Crafts):** Forms o-Chloro acetophenone and p-Chloro acetophenone.

(ii) **Nucleophilic Substitution Reaction of Chlorobenzene:**
Due to the resonance effect, the carbon-chlorine (C-Cl) bond in chlorobenzene has partial double bond character. This makes the C-Cl bond shorter and stronger than in alkyl halides, making it more difficult to break. As a result, chlorobenzene is much less reactive towards nucleophilic substitution reactions compared to alkyl halides. Nucleophilic substitution in chlorobenzene typically requires harsh conditions, such as high temperature and pressure, or the presence of strong electron-withdrawing groups on the benzene ring.
For example, Chlorobenzene can react with sodium hydroxide under high temperature (625 K) and high pressure to form phenol. Similarly, it can react with cuprous cyanide (CuCN) at high temperatures (475 K) to yield benzonitrile. Diphenyl can also be formed by reacting chlorobenzene with sodium in the presence of anhydrous ether.
In simple words: In chlorobenzene, the chlorine atom makes it harder for positive groups to attach, but when they do, they mostly go to specific spots (ortho and para). It's also hard for negative groups to replace the chlorine because that bond is strong due to special electron sharing.

🎯 Exam Tip: When explaining reactivity in chlorobenzene, always mention both the inductive (-I) and resonance (+M) effects of the halogen, and explain which effect dominates for specific reactions (e.g., +M for directing electrophiles, partial double bond for nucleophilic substitution).

 

Question 7. How will you prepare the following from alkyl halide:
(i) Alkyl isocyanide
(ii) Alkyl cyanide
(iii) Nitroalkane
(iv) Alkyl nitrite
(v) Isopropylbenzene
(vi) Tetramethylammonium chloride
Answer:
(i) **Alkyl isocyanide:** When an alkyl halide is treated with silver cyanide (\( \text{AgCN} \)), alkyl isocyanide is formed. This reaction occurs because the lone pair of electrons on the nitrogen atom of \( \text{AgCN} \) is more available for attack.
\( \text{R-X} + \text{AgCN} \xrightarrow {} \text{R-NC} + \text{AgX} \)
For example: \( \text{CH}_3\text{-Cl} + \text{AgCN} \rightarrow \text{CH}_3\text{-NC} + \text{AgCl} \)

(ii) **Alkyl cyanide:** Alkyl halides react with an alcoholic solution of potassium cyanide (\( \text{KCN} \)) to give alkyl cyanides. In this case, the carbon atom of \( \text{KCN} \) is more nucleophilic and attacks the alkyl group.
\( \text{R-X} + \text{KCN} \xrightarrow {} \text{R-CN} + \text{KX} \)
For example: \( \text{CH}_3\text{-Cl} + \text{KCN} \rightarrow \text{CH}_3\text{-CN} + \text{KCl} \)

(iii) **Nitroalkane:** When an alkyl halide is treated with silver nitrite (\( \text{AgNO}_2 \)), nitroalkanes are obtained. The nitrogen atom of \( \text{AgNO}_2 \) is the attacking site.
\( \text{R-X} + \text{AgNO}_2 \xrightarrow {} \text{R-NO}_2 + \text{AgX} \)
For example: \( \text{CH}_3\text{-CH}_2\text{-Br} + \text{AgNO}_2 \rightarrow \text{CH}_3\text{-CH}_2\text{-NO}_2 + \text{AgBr} \)

(iv) **Alkyl nitrite:** When an alkyl halide is treated with sodium nitrite (\( \text{NaNO}_2 \)) or potassium nitrite (\( \text{KNO}_2 \)), alkyl nitrite is obtained. Here, the oxygen atom of the nitrite group attacks the alkyl halide.
\( \text{R-X} + \text{NaNO}_2 \xrightarrow {} \text{R-ONO} + \text{NaX} \)
For example: \( \text{CH}_3\text{-CH}_2\text{-Br} + \text{NaNO}_2 \rightarrow \text{CH}_3\text{-CH}_2\text{-ONO} + \text{NaBr} \)

(v) **Isopropylbenzene:** Isopropylbenzene (cumene) is obtained when benzene reacts with isopropyl chloride (\( \text{CH}_3\text{-CH(Cl)-CH}_3 \)) in the presence of anhydrous aluminum chloride (\( \text{AlCl}_3 \)). This is a Friedel-Crafts alkylation reaction.
\( \text{C}_6\text{H}_6 + \text{CH}_3\text{-CH(Cl)-CH}_3 \xrightarrow { \text{Anhydrous AlCl}_3 } \text{C}_6\text{H}_5\text{-CH(CH}_3\text{)}_2 + \text{HCl} \)

(vi) **Tetramethylammonium chloride (Hoffmann Ammonolysis Reaction):** This reaction involves heating an alkyl halide (like chloromethane) with an alcoholic solution of ammonia in a sealed tube at about 373 K. This reaction doesn't stop at one product; instead, it forms a mixture of primary, secondary, and tertiary amines, and finally, a quaternary ammonium salt.
\( \text{CH}_3\text{-Cl} + \text{H-NH}_2 \xrightarrow { \text{C}_2\text{H}_5\text{OH}, \triangle } \text{CH}_3\text{-NH}_2 + \text{HCl} \) (Methylamine - Primary amine)
\( \text{CH}_3\text{-Cl} + \text{H-NHCH}_3 \xrightarrow { \text{C}_2\text{H}_5\text{OH}, \triangle } \text{CH}_3\text{-NH-CH}_3 + \text{HCl} \) (Dimethylamine - Secondary amine)
\( \text{CH}_3\text{-Cl} + \text{H-N(CH}_3\text{)}_2 \xrightarrow { \text{C}_2\text{H}_5\text{OH}, \triangle } \text{(CH}_3\text{)}_3\text{N} + \text{HCl} \) (Trimethylamine - Tertiary amine)
\( \text{(CH}_3\text{)}_3\text{N} + \text{CH}_3\text{-Cl} \xrightarrow {} \text{(CH}_3\text{)}_4\text{N}^+\text{Cl}^- \) (Tetramethyl ammonium chloride - Quaternary ammonium salt)
In simple words: Alkyl halides can make different products depending on what they react with. With silver cyanide, they make isocyanides; with potassium cyanide, they make cyanides. With silver nitrite, they make nitroalkanes; with sodium nitrite, they make alkyl nitrites. Isopropylbenzene is made by reacting benzene with isopropyl chloride. Finally, heating an alkyl halide with ammonia makes a mix of amines and a quaternary salt.

🎯 Exam Tip: Pay close attention to the specific reagents (e.g., \( \text{AgCN} \) vs. \( \text{KCN} \), \( \text{AgNO}_2 \) vs. \( \text{NaNO}_2 \)) as they determine whether the N or C atom (or N or O atom) attacks, leading to different products (isocyanide vs. cyanide, nitroalkane vs. alkyl nitrite).